\(q\)-torsion freeness of symmetric powers (Q1384099)
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scientific article; zbMATH DE number 1140075
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | \(q\)-torsion freeness of symmetric powers |
scientific article; zbMATH DE number 1140075 |
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\(q\)-torsion freeness of symmetric powers (English)
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13 May 1998
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Let \(R\) be a commutative noetherian unitary ring and \(E\) a finitely generated \(R\)-module. Let \(S_R(E)\) be the symmetric algebra of \(E\) over \(R\) and \(S_R(R^n)= R[X_1,\dots, X_n]\) the polynomial ring over \(R\). If \(E\) is of rank \(e\) and of finite projective dimension then there is an epimorphism between the cycles of the Koszul complex on the kernel of the canonical morphism \(S_R(R^n)\to S_R(E)\) and the cycles of the complex \(Z(E)\). Hence, if the Koszul complex \(K(f_1,\dots, f_m;S)\) of the ideal of relations of \(S(E)\) is exact then so is \(Z(E)\) (theorem 2.11.); here \(Z_r(E)= \text{Ker }\partial_r\) with \(\partial_R: \bigwedge^r R_1^n\to \bigwedge^{r-1} R^n\bigotimes_R E\) gives the complex \(Z(E)\): \[ 0\to Z_n(E)\otimes S[-n]\to Z_{n-1}(E)\otimes S[-n+1]\to \dots\to Z_0(E)\otimes S\to 0. \] For an integer \(q\) the module \(E\) is said to be \(q\)-torsion free if every \(R\)-regular sequence of length \(q\) is also \(E\)-regular. In the last part several results concerning the \(q\)-torsion freeness of the symmetric powers of \(E\) are proved.
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symmetric algebra
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Koszul complex
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\(q\)-torsion free module
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